-0.008 788 423 606 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 606 2(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.008 788 423 606 2(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.008 788 423 606 2| = 0.008 788 423 606 2


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.008 788 423 606 2.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.008 788 423 606 2 × 2 = 0 + 0.017 576 847 212 4;
  • 2) 0.017 576 847 212 4 × 2 = 0 + 0.035 153 694 424 8;
  • 3) 0.035 153 694 424 8 × 2 = 0 + 0.070 307 388 849 6;
  • 4) 0.070 307 388 849 6 × 2 = 0 + 0.140 614 777 699 2;
  • 5) 0.140 614 777 699 2 × 2 = 0 + 0.281 229 555 398 4;
  • 6) 0.281 229 555 398 4 × 2 = 0 + 0.562 459 110 796 8;
  • 7) 0.562 459 110 796 8 × 2 = 1 + 0.124 918 221 593 6;
  • 8) 0.124 918 221 593 6 × 2 = 0 + 0.249 836 443 187 2;
  • 9) 0.249 836 443 187 2 × 2 = 0 + 0.499 672 886 374 4;
  • 10) 0.499 672 886 374 4 × 2 = 0 + 0.999 345 772 748 8;
  • 11) 0.999 345 772 748 8 × 2 = 1 + 0.998 691 545 497 6;
  • 12) 0.998 691 545 497 6 × 2 = 1 + 0.997 383 090 995 2;
  • 13) 0.997 383 090 995 2 × 2 = 1 + 0.994 766 181 990 4;
  • 14) 0.994 766 181 990 4 × 2 = 1 + 0.989 532 363 980 8;
  • 15) 0.989 532 363 980 8 × 2 = 1 + 0.979 064 727 961 6;
  • 16) 0.979 064 727 961 6 × 2 = 1 + 0.958 129 455 923 2;
  • 17) 0.958 129 455 923 2 × 2 = 1 + 0.916 258 911 846 4;
  • 18) 0.916 258 911 846 4 × 2 = 1 + 0.832 517 823 692 8;
  • 19) 0.832 517 823 692 8 × 2 = 1 + 0.665 035 647 385 6;
  • 20) 0.665 035 647 385 6 × 2 = 1 + 0.330 071 294 771 2;
  • 21) 0.330 071 294 771 2 × 2 = 0 + 0.660 142 589 542 4;
  • 22) 0.660 142 589 542 4 × 2 = 1 + 0.320 285 179 084 8;
  • 23) 0.320 285 179 084 8 × 2 = 0 + 0.640 570 358 169 6;
  • 24) 0.640 570 358 169 6 × 2 = 1 + 0.281 140 716 339 2;
  • 25) 0.281 140 716 339 2 × 2 = 0 + 0.562 281 432 678 4;
  • 26) 0.562 281 432 678 4 × 2 = 1 + 0.124 562 865 356 8;
  • 27) 0.124 562 865 356 8 × 2 = 0 + 0.249 125 730 713 6;
  • 28) 0.249 125 730 713 6 × 2 = 0 + 0.498 251 461 427 2;
  • 29) 0.498 251 461 427 2 × 2 = 0 + 0.996 502 922 854 4;
  • 30) 0.996 502 922 854 4 × 2 = 1 + 0.993 005 845 708 8;
  • 31) 0.993 005 845 708 8 × 2 = 1 + 0.986 011 691 417 6;
  • 32) 0.986 011 691 417 6 × 2 = 1 + 0.972 023 382 835 2;
  • 33) 0.972 023 382 835 2 × 2 = 1 + 0.944 046 765 670 4;
  • 34) 0.944 046 765 670 4 × 2 = 1 + 0.888 093 531 340 8;
  • 35) 0.888 093 531 340 8 × 2 = 1 + 0.776 187 062 681 6;
  • 36) 0.776 187 062 681 6 × 2 = 1 + 0.552 374 125 363 2;
  • 37) 0.552 374 125 363 2 × 2 = 1 + 0.104 748 250 726 4;
  • 38) 0.104 748 250 726 4 × 2 = 0 + 0.209 496 501 452 8;
  • 39) 0.209 496 501 452 8 × 2 = 0 + 0.418 993 002 905 6;
  • 40) 0.418 993 002 905 6 × 2 = 0 + 0.837 986 005 811 2;
  • 41) 0.837 986 005 811 2 × 2 = 1 + 0.675 972 011 622 4;
  • 42) 0.675 972 011 622 4 × 2 = 1 + 0.351 944 023 244 8;
  • 43) 0.351 944 023 244 8 × 2 = 0 + 0.703 888 046 489 6;
  • 44) 0.703 888 046 489 6 × 2 = 1 + 0.407 776 092 979 2;
  • 45) 0.407 776 092 979 2 × 2 = 0 + 0.815 552 185 958 4;
  • 46) 0.815 552 185 958 4 × 2 = 1 + 0.631 104 371 916 8;
  • 47) 0.631 104 371 916 8 × 2 = 1 + 0.262 208 743 833 6;
  • 48) 0.262 208 743 833 6 × 2 = 0 + 0.524 417 487 667 2;
  • 49) 0.524 417 487 667 2 × 2 = 1 + 0.048 834 975 334 4;
  • 50) 0.048 834 975 334 4 × 2 = 0 + 0.097 669 950 668 8;
  • 51) 0.097 669 950 668 8 × 2 = 0 + 0.195 339 901 337 6;
  • 52) 0.195 339 901 337 6 × 2 = 0 + 0.390 679 802 675 2;
  • 53) 0.390 679 802 675 2 × 2 = 0 + 0.781 359 605 350 4;
  • 54) 0.781 359 605 350 4 × 2 = 1 + 0.562 719 210 700 8;
  • 55) 0.562 719 210 700 8 × 2 = 1 + 0.125 438 421 401 6;
  • 56) 0.125 438 421 401 6 × 2 = 0 + 0.250 876 842 803 2;
  • 57) 0.250 876 842 803 2 × 2 = 0 + 0.501 753 685 606 4;
  • 58) 0.501 753 685 606 4 × 2 = 1 + 0.003 507 371 212 8;
  • 59) 0.003 507 371 212 8 × 2 = 0 + 0.007 014 742 425 6;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.008 788 423 606 2(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1000 1101 0110 1000 0110 010(2)

6. Positive number before normalization:

0.008 788 423 606 2(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1000 1101 0110 1000 0110 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.008 788 423 606 2(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1000 1101 0110 1000 0110 010(2) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1000 1101 0110 1000 0110 010(2) × 20 =

1.0001 1111 1111 1010 1010 0011 1111 1100 0110 1011 0100 0011 0010(2) × 2-7


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -7

Mantissa (not normalized):
1.0001 1111 1111 1010 1010 0011 1111 1100 0110 1011 0100 0011 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1016(10) =

011 1111 1000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 1111 1111 1010 1010 0011 1111 1100 0110 1011 0100 0011 0010 =

0001 1111 1111 1010 1010 0011 1111 1100 0110 1011 0100 0011 0010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1000

Mantissa (52 bits) =
0001 1111 1111 1010 1010 0011 1111 1100 0110 1011 0100 0011 0010


Decimal number -0.008 788 423 606 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0011 1111 1100 0110 1011 0100 0011 0010

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100